I've been trying to solve the inhomogeneous PDE IVP $u_t+c u_x = e^{2x}$; $u(x,0)=f(x)$, but got stuck. Would appreciate some help.

Here's what I did by trying to use the method of characteristics:

$\frac{dt}{ds}=1$, $\frac{dx}{ds}=c$, $\frac{dz}{ds}=e^{2x}$, where $s$ is the parametrization variable for a characteristic curve lying in the surface $u(x,t)$.

So I get $t(s) = s+C_1$, $x(s) = cs+C_2$, $z(s) = e^{2x}+C_3$. We can thus see that $ct-x = C_1-C_2 = D_1$, where $C_i$ and $D_i$ are some constants.

But here's where I'm stuck since I have no idea what should be done next. That is, how can we now relate $ct-x$ to $z(x,t)$ to build the surface?

I was thinking that, maybe, do something like this: $z-te^{2x}=C_3-C_1=D_2$, so $z(x,t) = D_2 + te^{2(ct-D_1)}$ (after substituting). But is this the solution then and what's remaining is to apply the initial condition?

$z(x,t) = u(x,t)$, $u(x,0) = D_2 = f(x)$, so $u(x,t) = f(x) + te^{2(ct-D_1)}$. Unfortunately, this does not match the solution from WolframAlpha.

  • $\begingroup$ Firstly, you integrated $dz/ds$ incorrectly. Secondly, I always find it is easier to just eliminate the $s$ variable using the fact that $dt/ds = 1 \implies dt = ds$. So your ODE solution for $x$ becomes $$x(t) = ct + x_{0}$$ and your other ODE becomes $$dz/dt = e^{2x} = e^{2ct + 2x_{0}} = x_{0}e^{2ct} \implies \cdots$$ $\endgroup$ – mattos May 26 '16 at 1:02
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    $\begingroup$ After solving for $x(s) = cs + x_0$ we use this in the equation for $u$ (which you call $z$) to get $\frac{du}{ds} = e^{2x} = e^{2(cs + x_0)} \implies u(s) = u_0 + \frac{1}{2c}e^{2x(s)}$ thus the solution is $u(x,t) = \frac{1}{2c}e^{2x} + g(x - ct)$ where $g$ is determined by imposing the initial conditions. Another option to solve it is to eliminate the source first (since it's a simple function of $x$) by taking $u(x,t) = v(x,t) + h(x)$. This gives $[v_t + cv_x] + [ch'(x) - e^{2x}] = 0$. Taking $h(x)$ such that $ch'(x) - e^{2x} = 0$ we get the simpler PDE $v_t + cv_x = 0$. $\endgroup$ – Winther May 26 '16 at 1:17
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    $\begingroup$ I have no idea why it chooses that form - that depends on how they have implemented the algorithm. As I said it does not matter how you write it as $C_1$ is a free function. Try to convince yourself that the three ways I have written the solution above (the WA, the $g$ way and the $h$ way) are equvialent. If you are in doubt try to impose the initial condition for each of the cases and see that you end up with the same result. $\endgroup$ – Winther May 26 '16 at 1:41
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    $\begingroup$ @RodrigodeAzevedo One can't use separation of variables directly when the right hand side is not zero. The other problem with separation of variables is that the expression for the final solution will be much more complicated (involving an infinite series where the coefficients are given as integrals over $f(x)$). With the method of charactersistics one gets the a simple expression for the solution $f(x-ct) + \frac{1}{2c}[e^{2x}-e^{2x-2ct}]$ which is easy to evaluate for any $f(x)$. $\endgroup$ – Winther May 26 '16 at 1:44
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    $\begingroup$ @RodrigodeAzevedo To add to the comment above. If we apply separation of variables (after having removed the source) then we find that since this is not a boundary value problem the separation constants are completely free. I guess this means one could write the solution as $$u(x,t) = \int g(\lambda)e^{\lambda(x-ct)}{\rm d}\lambda + \frac{1}{2c}e^{2x}$$ where $g$ is some free function. The integral should be equivalent to a free function of $x-ct$ so I guess the method could work here. Anyway the method of characteristics is a much cleaner way of deriving the solution. $\endgroup$ – Winther May 26 '16 at 2:03

The answer to the question raised by "sequence" is given in the comments. So, my answer is only a different form (but equivalent) of the method of characteristics with advantage of clearness :

$$u_t+cu_x=e^{2x}$$ The characteristic equations are : $\quad \frac{dt}{1}=\frac{dx}{c}=\frac{du}{e^{2x}}$

From $\frac{dt}{1}=\frac{dx}{c}$ , the first characteristic curve : $$x-ct=c_1$$

From $\frac{dx}{c}=\frac{du}{e^{2x}} \quad\to\quad du-\frac{1}{c}e^{2x}dx=0$ , the second characteristic curve : $$u-\frac{1}{2c}e^{2x} =c_2$$

Thus, the general solution of the PDE, expressed on implicit form, is : $$\Phi\left((x-ct)\:,\:(u-\frac{1}{2c}e^{2x})\right)=0$$ where $\Phi$ is any differentiable function of two variables.

Solving for the second variable leads to the explicit form : $\quad (u-\frac{1}{2c}e^{2x})=F(x-ct)$ $$u=\frac{1}{2c}e^{2x}+F(x-ct)$$ where $F$ is any differentiable function.

The condition : $u(x,0)=f(x)$ implies $\frac{1}{2c}e^{2x}+F(x-0)=f(x) \quad\to\quad F(x)=f(x)-\frac{1}{2c}e^{2x}$

$ F(x-ct)=f(x-ct)-\frac{1}{2c}e^{2(x-ct)}$



  • $\begingroup$ How do you come up with $\Phi\left((x-ct)\:,\:(u-\frac{1}{2c}e^{2x})\right)=0$? $\endgroup$ – sequence May 26 '16 at 20:37
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    $\begingroup$ This is a basic in the method of characteristics. For example, see the theorem 2.5, page 4 in : math.ualberta.ca/~xinweiyu/436.A1.12f/… $\endgroup$ – JJacquelin May 27 '16 at 4:13

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